Selçuk J. Appl. Math. Selçuk Journal of Vol. 11. No.1. pp. 81-94 , 2010 Applied Mathematics
Age and Block Replacement Policies in Renewal Processes Duygu Sava¸sc¬1,Halil Aydo¼gdu2
1Georg-August-Universität Göttingen Institut für Statistik und Ökonometrie, Platz
der Göttinger Sieben 5 37073 Göttingen, Germany e-mail: D uygu.Savasci@ w iw i.uni-go ettingen.de
2Department of Statistics Faculty of Science, Ankara University 06100 Tando¼
gan-Ankara, Türkiye
e-mail: aydogdu@ science.ankara.edu.tr
Received Date: March 24, 2009 Accepted Date: September 4, 2009
Abstract. In this study, renewal processes, age and block replacement policies are described. Certain classes of distribution functions which are related to the usage of these policies are given. Some results on age and block replacement policies in renewal processes are reviewed.
Key words: Age replacement policy; block replacement policy; renewal theory; stochastic order; Laplace order.
2000 Mathematics Subject Classi…cation: 60K05. 1. Introduction
In many situations, failure of a unit during actual operation is costly or danger-ous. Replacement policies are followed to reduce the incidence of system failure. Some useful replacement policies in popular use are the age replacement policy and the block replacement policy. Under an age replacement policy, a unit is replaced upon failure or at age T, a speci…ed positive constant, whichever comes …rst. Under block replacement policy, a unit is replaced upon a failure and at times T, 2T, 3T,...
Age replacement is administratively more di¢ cult to implement, since the age of the unit must be recorded. Block replacement is simpler to administer since the age of the unit need not to be recorded. It leads to more frequent replacement of relatively new items. This type of policy is commonly used with computers and other complex electronic systems. On the other hand, age replacement is more ‡exible since planned replacement takes into account the age of the unit under
this policy. Therefore, it is some of interest to compare these two policies with respect to the number of failures, the number of planned replacements, and the number of removals. Age and block replacement policies have been investigated by Barlow and Proschan (1975) and Yue and Cao (2001), among others. All the replacements can be treated by the techniques of renewal theory. Before investigating the replacement problem, it is necessary to present some classes of life distributions and the renewal theory.
2. Some Classes of Life Distributions
Assume that the life length of a unit has a distribution function F with F (x) = 0 for x < 0. The corresponding survival function is denoted by F = 1 F . Increasing (Decreasing) Failure Rate
A distribution function F is IFR (DFR) if for all x > 0, F (x+t)=F (t) is decreas-ing (increasdecreas-ing) in t whenever t 0 and F (t) > 0. If F has a density function f , then this is equivalent to the condition that the failure rate r(t) = f (t)=F (t) is increasing (decreasing) in t on t : F (t) > 0 .
New Better (Worse) Than Used
A distribution function F is NBU (NWU) if
(1) F (x + y) ( )F (x) F (y)
for x 0; y 0. Equality in (1) holds if and only if F is the exponential distribution function.
New Better (Worse) Than Used in Expectation A distribution function F is NBUE (NWUE) if
i) F has …nite (or in…nite mean) , ii)Rt1F (x) dx ( ) F (t) for t 0
where Rt1[ F (x)=F (t) ] dx represents the conditional mean remaining life of a unit of age t.
It is well known that
IF R ) NBU ) NBUE; and
New Better (Worse) Than Used in Laplace Ordering
Let X and Y be two nonnegative random variables with distribution functions F and G, respectively. It is said that X is smaller than Y (or F is smaller than G) with respect to Laplace order L, denoted by X LY (F LG) if
Z 1 0 e stdF (t) Z 1 0 e stdG(t); s 0:
Let Xtdenote the residual life of X at age t 0. The nonnegative random vari-able X is said to be NBUL(NWUL) if and only if Xt L( L)X or equivalently
Z 1 0 e sxF (t + x)dx ( ) F (t) Z 1 0 e sxF (x)dx: Wang (1996) has shown that the following expressions hold;
N BU ) NBUL ) NBUE; N W U ) NW UL ) NW UE:
3. Renewal Theory
A renewal process is a sequence of independent, identically distributed, nonneg-ative random variables X1; X2; :::, which, with probability 1, are not all zero. Let F be the distribution function of X1; F is called the underlying distribution function of the renewal process. Fk , the k-fold Stieltjes convolution of F with itself, is the distribution function of Sk X1+ X2+ ::: + Xk.
Renewal theory is primarily concerned with the number of renewals N (t) in [0; t]. N (t), the renewal random variable is the maximum value of k for which Sk t. The stochastic process fN(t); t 0g is also known as a renewal count-ing process (Ross 1983).
It is known that P (N (t) n) = P (Sn t) = Fn (t) for n = 0; 1; ::: . It follows that P (N (t) = n) = P (N (t) n) P (N (t) n + 1), so that
P (N (t) = n) = Fn (t) F(n+1) (t) ; n = 0; 1; ::: :
The mean value function (renewal function) of the renewal process fN(t); t 0g is; M (t) = E(N (t)) = 1 X k=1 P (N (t) k) = 1 X k=1 Fk (t) ; t 0:
It is well known that the renewal function M (t) satis…es the integral equation (renewal equation); (2) M (t) = F (t) + Z t 0 M (t x)dF (x); t 0: If F has a density f , di¤erentiation of (2) yields
m(t) = f (t) + Z t
0
m(t x)f (x)dx; t 0; where m(t) = dtdM (t) is known as the renewal density.
Note that N (t) is the number of renewals in [0; t], and the next arrival will be that numbered N (t) + 1. That is to say, we have begun our observation at a point in the random interval It = [SN (t); SN (t)+1), the endpoints of which are arrival times. E(t) = SN (t)+1 t is called the excess lifetime at t and C(t) = t SN (t) is called the current lifetime (or age) at t (Grimmett and Stirzaker 1992). That is, E(t) is the time which elapses before the next arrival. C(t) is the elapsed time since the last arrival. It is known that P (C(t) t) = 1 and P (C(t) = t) = 1 F (t). Since E(t y) > y if and only if no arrivals occur in (t y; t], P (C(t) y) = P (E(t y) y), y < t.
Renewal Reward Processes
Consider a renewal process fN(t); t 0g having interarrival times Xn; n 1 with distribution function F , and suppose that each time a renewal occurs we receive a reward. Let Rn denote the reward earned at the time of the nth renewal. It is assumed that the Rn; n 1, are independent and identically distributed. However, it is allowed for the possibility that Rn may depend on Xn, the length of the nth renewal interval, and so it is assumed that the pairs (Xn; Rn); n 1, are independent and identically distributed. We consider fR(t); t 0g where R(t) =PN (t)n=1 Rn. Then, R(t) represents the total reward earned by time t. If E(jRj) < 1 and E(X) < 1, then as t ! 1,
i)R(t)=t ! E(R)=E(X) with probability 1, ii)E(R(t))=t ! E(R)=E(X)
where E(R) = E(Rn) and E(X) = E(Xn) (Ross 1983). This theorem is known in the literature as renewal-reward theorem.
4. Age and Block Replacement Policies
We will assume for both policies that units fail permanently, independently and that the time required to perform replacement is negligibly small.
Age Replacement Policy
Under an age replacement policy, a unit is replaced upon failure or at age T, whichever comes …rst. Denote the number of failures in [0; t] by N (t) and let fXig1i=1 represent the durations between successive failures with distribution function F . Denote the total number of removals in [0; t] by NA(t; T ) under age replacement policy with replacement interval T , and denote the number of failures in [0; t] by NA0(t; T ) under age replacement policy with replacement interval T .
Theorem 4.1. (Karlin and Taylor 1975) fNA(t; T ); t 0g is a renewal process with interrenewal time distribution function
FA(x) = F (x); x < T
1; x T
and the mean renewal duration isR0T(1 F (x)) dx.
Proof: Let fYig1i=1 denote the durations between successive removals. Then, Yi = minfXi; T g for i = 1; 2; ::: . Since Xi, i = 1; 2; :::, are independent identically distributed, Yi, i = 1; 2; :::, are independent identically distributed as well. Thus, fNA(t; T ); t 0g is a renewal process. It is clear that (Y1 > y) (X1> y; T > y). Therefore, P (Y1> y) = F (y); T > y 0; T y is acquired. Hence, FA(x) = F (x); x < T 1; x T Also, E(Y1) = R1 0 (1 FA(y)) dy = RT 0 (1 F (x)) dx.
Theorem 4.2. (Barlow and Proschan 1975) fNA0(t; T ); t 0g is a renewal process with interrenewal time distribution function,
GA(x) = 1 F (T )nF (x nT ); nT x (n + 1)T; n = 0; 1; ::: and the renewal duration has the expectation F (T )1 R0T(1 F (x)) dx.
Proof: Let fYig1i=1 denote the durations between unplanned failures under age replacement policy. According to the process, from the time that an un-planned failure occurs, the process continues probabilistically in the same way. Therefore, fYig1i=1 random variables are independent identically distributed. P (Y1> x) = P (X1> x) = F (x), for 0 < x < T ,
P (Y1> x) = P (X1> T; X2> x T ) = F (T ) F (x T ), for T < x < 2T , P (Y1 > x) = P (X1 > T; X2 > T; X3 > x 2T ) = F (T ) F (T ) F (x 2T ), for 2T < x < 3T .
In general it can be written as
P (Y1> x) = F (T )nF (x nT ); nT x (n + 1)T; n = 0; 1; ::: It follows that
GA(x) = 1 F (T )nF (x nT ); nT x (n + 1)T; n = 0; 1; ::: Also, the expectation of Y1 is
E(Y1) = Z T 0 F (x) dx + Z 2T T F (T )F (x T ) dx + Z 3T 2T F (T )2F (x 2T ) dx + ::: = 1 X n=0 Z (n+1)T nT F (T )nF (x nT ) dx = 1 X n=0 F (T )n Z T 0 F (y) dy = 1 F (T ) Z T 0 F (y) dy :
An Optimal Policy for Age Replacement
We assume an age replacement which is basic for controlling items that are subject to stochastic breakdowns. A cost of cp> 0 is incurred for each planned replacement and a cost of cf for each failure replacement where cf > cp. Under age replacement policy, it is obvious that the cost incurred during one replace-ment cycle is the random variable Y which is
Y = cf; X1< T cp; X1 T: E(Y ) = cfF (T ) + cp(1 F (T )): Hence, by the renewal-reward theorem,
the long run average cost per unit time = cpR+ (cT f cp)F (T ) 0 (1 F (x)) dx
with probability 1. By putting the derivative of the average cost function equal to zero, it is veri…ed that the minimizing value of T is the unique solution to the equation r(T ) Z T 0 (1 F (x)) dx F (T ) = cp cf cp
where r(T ) is the failure rate function de…ned by r(T ) = f (T )=F (T ), and it is assumed that this function is continuous and strictly increasing to in…nity (Tijms 1995).
Denote the long-run average cost per unit time for the age replacement rule with limit T by g(T ) and let T be the optimal value of T . Then,
g(T ) = cpR+ (cT f cpF (T )) 0 (1 F (x)) dx and r(T ) Z T 0 (1 F (x)) dx F (T ) = cp cf cp : Thus, g(T ) = cpc+ (cf cp)F (T ) p cf cp 1 r(T )+ F (T ) r(T ) = (cf cp) r(T ):
Block Replacement Policy
Under a block replacement policy, a unit is replaced by a new one upon failure and upon scheduled times T; 2T; ::: . There is always a replacement at the sched-uled times regardless of the age of the item in use. Denote the total number of removals in [0; t] by NB(t; T ) under block replacement policy with replacement interval T , and denote the number of failures in [0; t] by NB0 (t; T ) under block replacement policy with replacement interval T .
While fNA(t; T ); t 0g and fN
0
A(t; T ); t 0g are renewal processes, fNB(t; T ); t 0g is not a renewal process.
Let the times between removals be fYig1i=1 in fNB(t; T ); t 0g. Then, Y1= minfX1; T g
Y2= minfX2; T g; Y1= T minfX2; T X1g; Y1= X1: It is easily obtained that
FY1(y) =
F (y); y < T
1; y T
and
FY2(y) =
F (y) + (1 F (y))(F (T ) F (T y)); y < T
As we see from the distribution functions of Y1and Y2, they are not identically distributed. Hence, fNB(t; T ); t 0g process cannot be a renewal process. Furthermore, the counting process fNB0 (t; T ); t 0g is also not a renewal process.
´
An Optimal Policy for Block Replacement
Assume that the cost structure is the same as in the age replacement policy. The stochastic process describing the age of the item in use is regenerative. The length of one cycle is T . Further,
E(Y ) = cp+ cfM (T )
where the random variable Y is the cost that is incurred during one replacement cycle and M denotes the renewal function associated with the lifetime distri-bution function F . This follows by noting that the number of renewals up to time T in the renewal process generated by the lifetimes X1; X2; ::: is nothing else than the number of failure replacements up to time T . Hence, for the block replacement with parameter T ,
the long run average cost per unit time = 1
T fcp+ cfM (T )g with probability 1 (Tijms 1995). If the T value which makes this function min-imum exists, then the policy that we use with this T value is the optimal policy for the block replacement.
Replacement Comparisons
It is useful to compare block replacement with age replacement, using replace-ment interval T for both of them. For example, block replacereplace-ment is more wasteful since more unfailed components are removed than under age replace-ment. Under the IFR assumption, the expected number of failures will be less under block replacement. The following theorem, true for all distributions, is intuitively obvious.
Theorem 4.3. P (N (t) n) P (NA(t; T ) n) P (NB(t; T ) n) for all t 0, n = 0; 1; ::: .
Proof: Let fXig1i=1 represent the durations between successive failures. Let SA
n (SnB) denote the time of the nth removal under age (block) replacement pol-icy.
YB
1 = minfX1; T g, YB
Y3B= minfX3; 3g, 0 3 T , generally, we have Yn
B = minfXn; ng, for 0 n T under block replacement policy, where n is the remaining life to a planned renewal after (n 1)th removal. Under age replacement policy; YA
n = minfXn; T g for n = 1; 2; ::: . Then,
SnA= Sn 1A + minfXn; T g; SnB= Sn 1B + minfXn; ng: Since initially SA
1 = S1B; Sn SnA SnB. Thus, the proof is completed (Barlow and Proschan 1963).
Let X and Y be any two random variables. It is said that the random vari-able X is stochastically larger than the random varivari-able Y , written X st Y , if P (X > a) P (Y > a) for all a. Also from Theorem 4.3 we have N (t) st NA(t; T ) stNB(t; T ) which means that the number of removals in [0; t] inter-val under block replacement policy is stochastically bigger than the number of removals under age replacement policy.
Corollary: M (t) E(NA(t; T )) E(NB(t; T )).
This corollary is an immediate consequence of Theorem 4.3. It is shown by Barlow and Proschan (1963) that if F is IFR, then
P (N (t) n) P (NA0(t; T ) n) P (NB0(t; T ) n) for t 0 and n = 0; 1; ::: which means N (t) stN
0
A(t; T ) stN
0
B(t; T ). Equality is attained for the exponential distribution F (x) = 1 e x= 1 where
1denotes the mean of F . As a consequence of this, we have M (t) E(NA0(t; T )) E(NB0 (t; T )).
Theorem 4.4. N (t) st N
0
A(t; T ) for all t 0, T 0 , F is NBU (Barlow and Proschan, 1975).
Theorem 4.4 states that the class of NBU distributions is the largest class for which age replacement diminishes stochastically the number of failures expe-rienced in any particular time interval [0; t]. In this sense, the NBU class of distributions is a natural class to consider in age replacement.
Let F be IFR. For …xed T > 0, we know that fNA0 (t; T ); t 0g is a renewal process with underlying distribution function FA(t; T ) = 1 F (T )nF (x nT ), where nT x (n + 1)T for n = 0; 1; ::: . By di¤erentiating FA(t; T ) with respect to T , it is veri…ed that FA(t; T ) is increasing in T 0 for …xed t 0. Hence P (NA0(t; T ) n) = FAn (t; T ) is increasing in T 0 for …xed t 0, where FAn (t; T ) is the n-fold Stieltjes convolution of FA(t; T ) with itself (Bar-low and Proschan 1975). Thus, under age replacement policy with an IFR failure
distribution, the number of failures observed in any interval [0; t] increases sto-chastically as the replacement interval T increases. Also, the direct contrary is true (Marshall and Proschan 1972).
Lemma 4.1. Consider two policies such that the planned replacements oc-cur at …xed time points ft1; t2; :::g under policy 1, and at the time points ft1; t2; :::g [ ft0g under policy 2, where 0 < t1 < t2; ::: . Let Ni(t) be the number of failures in [0; t] under policy i, i = 1; 2. Then N1(t) st N2(t) for each t 0 if and only if the underlying life distribution function F is NBU (Barlow and Proschan 1975).
Theorem 4.5. N (t) st N
0
B(t; T ) for all t 0, T 0 , F is NBU (Barlow and Proschan 1975).
Theorem 4.5 states that the class of NBU distributions is the largest class for which block replacement diminishes stochastically the number of failures in any particular time interval [0; t], 0 < t < 1.
We know that under age replacement policy, the number of failures observed in any interval [0; t] increases stochastically as the replacement interval T increases if and only if F is IFR. Under block replacement policy, it is shown by Shaked and Zhu (1992) that the stochastic increasingness of NB0(t; T ) in T 0, for each …xed t 0, is a su¢ cient condition for F to be IFR, but it is not a necessary condition. Suppose that fNB0(t; T )g increases stochastically as the replacement interval T increases for …xed t 0. Hence, P (NB0(t; T2) n) P (N
0
B(t; T1) n) for T1 T2. We choose n = 1, then it is clear that
P (NB0(t; T2) = 0) = 1 P (N
0
B(t; T2) 1) 1 P (NB0 (t; T1) 1) = P (NB0(t; T1) = 0):
Let T1 T2 2T1 and we choose t such that T1 T2 t 2T1. Since P (NB0(t; T1) = 0) = F (T1) F (t T1) and P (N
0
B(t; T2) = 0) = F (T2) F (t T2), we have
(3) F (T1) F (t T1) F (T2) F (t T2):
We need to show F (x + )=F (x) F (y + )=F (y) for any given y x 0 and 0.
i)Let > y x. Then, we take T1= x + , T2= y + and t = x + y + . Then, T1, T2 and t hold 0 T1 T2 t 2T1. Since t T1= y, t T2= x, T1= x + and T2= y + , by (3) we have F (x + )=F (x) F (y + )=F (y). ii) Let y x. Then, we take T1= y, T2 = y + and t = x + y + . Then, T1, T2and t hold 0 T1 T2 t 2T1. Since t T1= x+ , t T2= x, T1= y and T2= y + , by (3) we have F (x + )=F (x) F (y + )=F (y).
Hence, if NB0(t; T ) "st in T 0 for each …xed t 0, then F is IFR.
Now, some comparisons are given for age and block replacement policies under the assumption that the underlying distribution function F is NBUL or NWUL due to Yue and Cao (2001).
Let fXig1i=1and fYig1i=1denote the interrenewal times for the renewal processes fN(t); t 0g andnNA0(t); t 0o. It is obvious that
X1 L( L)Y1, Xi L( L)Yi i = 1; 2; ::: : Theorem 4.6. F is NBUL(NWUL) , X1 L ( L)Y1.
Proof: We know from Theorem 4.2 that P (Y1> x) = F (T )nF (x nT ); nT x (n + 1)T; n = 0; 1; ::: . Thus, Z 1 0 e sxP (Y1> x) dx = 1 X n=0 Z (n+1)T nT e sxF (T )nF (x nT ) dx = 1 X n=0 [F (T ) e sT]n Z T 0 e sxF (x) dx = Z T 0 e sxF (x) dx =[1 e sTF (T )]: From the de…nition of Laplace ordering we have
X1 L( L)Y1, Z 1 0 e sxF (x) dx ( ) Z 1 0 e stP (Y1> x) dx: Hence, X1 L ( L)Y1, Z 1 0 e sxF (x) dx ( ) Z T 0 e sxF (x) dx =[1 e sTF (T )]: This is equivalent to F is NBUL(NWUL) (Yue and Cao 2001).
Let fN1(t); t 0g and fN2(t); t 0g be two counting processes such that Z 1 0 e stP (N1(t) n) dt Z 1 0 e stP (N2(t) n) dt
for all s > 0 and n = 0; 1; :::. Then, fN1(t); t 0g is said to be smaller than fN2(t); t 0g in Laplace order and denoted as N1(t) LN2(t).
Theorem 4.7. Let fN1(t); t 0g and fN2(t); t 0g be two renewal processes. Xi1and Xi2denote the duration between (i 1)th and ith renewal for fN1(t); t 0g and fN2(t); t 0g, respectively. Then, N2(t) L N1(t) if and only if
Xi1 LXi2, i = 1; 2; ::: . Proof: \ ) \ : Z 1 0 e stP (N1(t) 1) dt Z 1 0 e stP (N2(t) 1) dt or equivalently Z 1 0 e stP (N1(t) = 0) dt Z 1 0 e stP (N2(t) = 0) dt :
Observing that P (Nk(t) = 0) = P (Xik > t) for k = 1; 2 and i = 1; 2; ::: we have Z 1 0 e stP (Xi1> t) dt Z 1 0 e stP (Xi2> t) dt : Thus, X1 i LXi2, i = 1; 2; ::: . \ ( \ : We have n P i=1 X1 i L n P i=1 X2
i (Alzaid, Kim and Proschan 1991), or equivalently Z 1 0 e stP ( n X i=1 Xi1> t) dt Z 1 0 e stP ( n X i=1 Xi2> t) dt: Hence, Z 1 0 e stP ( n X i=1 Xi1 t) dt Z 1 0 e stP ( n X i=1 Xi2 t) dt: It follows that Z 1 0 e stP (N1(t) n) dt Z 1 0 e stP (N2(t) n) dt for all s > 0 and n = 1; 2; ::: . Thus, N2(t) LN1(t) (Yue and Cao 2001). As a consequence of Theorem 4.6 and Theorem 4.7 we have;
N (t) L( L)N
0
A(t; T ) , F is NBUL(NWUL),
which states that the age replacement diminishes (increases) , in the sense of Laplace order, the number of failures in any particular time interval [0; t], 0 < t < 1, if and only if F is NBUL(NWUL).
Lemma 4.2. Let planned replacements occur at …xed time points f0 < t1 < t2 < :::g under policy 1, and at time points f0 < t1 < t2 < :::g [ ft0g under policy 2. Let Ni(t) be the number of failures in [0; t] under policy i, i = 1; 2.
Then, N1(t) L ( L)N2(t) for each t 0 if and only if the underlying life distribution F is NBUL(NWUL).
The proof of Lemma 4.2 is similar to that of Lemma 4.1. The next theorem given by Yue and Cao (2001) states that the block replacement diminishes (in-creases), in the sense of Laplace order, the number of failures experienced in any particular time interval [0; t], 0 < t < 1, if and only if F is NBUL(NWUL). Theorem 4.8. N (t) L( L)N
0
B(t; T ), t 0, T 0 , F is NBUL(NWUL). Proof: \ ( \ : Let planned replacements occur at …xed time points f0; T; :::; (i 1)T g under policy i, i = 1; 2; ::: . Let Ni(t) be the number of failures in [0; t] under policy i, i = 1; 2; ::: . It follows from Lemma 4.2 that N (t) = N1(t) L N2(t) L::: L Nk(t) L::: . As k ! 1, we have N(t) LN 0 B(t; T ). \ ) \ : It is clear that Z 1 0 e stP (N (t) 1) dt Z 1 0 e stP (NB0(t; T ) 1) dt: Hence, (4) Z 1 0 e stP (N (t) = 0) dt Z 1 0 e stP (NB0(t; T ) = 0) dt: Observing that P (NB0 (t; T ) = 0) = P (X1> T; :::; Xk> T; Xk+1> t kT ) = F (T )kF (t kT ) for kT t < (k + 1)T , k = 0; 1; ::: . Then, Z 1 0 e stP (NB0 (t; T ) = 0) dt = 1 X k=0 Z (k+1)T kT e stF (T )kF (t kT ) dt = 1 X k=0 [F (T ) e sT]k Z T 0 e stF (t) dt = Z T 0 e stF (t) dt=(1 e sTF (T )):
Noting thatR01 e stP (N (t) = 0) dt =R01e stF (t) dt and from (4) we obtain Z 1 0 e stF (t) dt RT 0 e stF (t) dt 1 e sTF (T ): Thus, F is NBUL. The proof in NWUL case is similar.
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